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F.4 數學 Number system 一條
發問:
If z is a complex number and (1+i)z^2-(2+3i)+2+i=0. Find z. 請問如何計算？ 更新: 題目沒有打錯，謝謝！ 更新 2: 已改為上傳圖片： http://img822.imageshack.us/img822/5615/123ou.png 謝謝！ 更新 3: 答案應該不一定是數字，因為上個幾題有ｄanswer都係complex number
最佳解答:
If z is a complex number and (1+i)z^2－(2+3i)+2+i=0. Find z.請問如何計算？Sol(1+i)z^2－(2+3i)+2+i=0(1+i)z^2=2+3i－2－i(1+i)z^2=2iz^2=2i/(1+i)=2(1－i)i/2=i(1－i)=1+i=√2(1/√2+i/√2) =2^(1/2)(Cos(π/4)+iSin(π/4)) =2^(1/2)(Cos(2nπ+π/4)+iSin(2nπ+π/4)) z=2^(1/4)(Cos(nπ+π/8)+iSin(nπ+π/8)) (1) n=0z=2^(1/4) (Cos(π/8)+iSin(π/8)) =2^(1/4) (√(2+√2)/2+i√(2－√2)/2) =2^(－3/4) (√(2+√2)+i√(2－√2)) (2) n=1z=2^(1/4) (Cos(9π/8)+iSin(9π/8)) =2^(1/4) (－√(2+√2)/2－i√(2－√2)/2) =2^(－3/4) (－√(2+√2)－i√(2－√2)) 題目改為(1+i)z^2－(2+3i)z+2+i=0. (1+i)z^2－(2+3i)z+2+i=0D=(2+3i)^2－4*(1+i)(2+i) =(4+12i－9)－4(2+i+2i－1) =(4+12i－9)－4(1+3i) =－9(1+i)z^2－(2+3i)z+2+i=0z=((2+3i)+/-3i)/(2+2i) (1) z=(2+3i+3i)/(2+2i) =(1+3i)/(1+i) =(1+3i)(1－i)/[(1+i)(1－i)] =(1－i+3i+3)/2=2+i(2) z=((2+3i)－3i)/(2+2i) =2/(2+2i) =1/(1+i) =(1－i)/2 2010-09-03 00:15:19 補充： If z is a complex number and (1+i)z^2－(2+3i)+2+i=0. Find z. 請問如何計算？ Sol (1+i)z^2－(2+3i)+2+i=0 (1+i)z^2=2+3i－2－i (1+i)z^2=2i z^2=2i/(1+i)=2(1－i)i/2=i(1－i)=1+i =√2(1/√2+i/√2) =2^(1/2)(Cos(π/4)+iSin(π/4)) =2^(1/2)(Cos(2nπ+π/4)+iSin(2nπ+π/4)) z=2^(1/4)(Cos(nπ+π/8)+iSin(nπ+π/8)) 2010-09-03 00:16:16 補充： (1) n=0 z=2^(1/4) (Cos(π/8)+iSin(π/8)) =2^(1/4) (√(2+√2)/2+i√(2－√2)/2) =2^(－3/4) (√(2+√2)+i√(2－√2)) (2) n=1 z=2^(1/4) (Cos(9π/8)+iSin(9π/8)) =2^(1/4) (－√(2+√2)/2－i√(2－√2)/2) =2^(－3/4) (－√(2+√2)－i√(2－√2)) 2010-09-03 00:16:53 補充： 題目改為(1+i)z^2－(2+3i)z+2+i=0. (1+i)z^2－(2+3i)z+2+i=0 D=(2+3i)^2－4*(1+i)(2+i) =(4+12i－9)－4(2+i+2i－1) =(4+12i－9)－4(1+3i) =－9 2010-09-03 00:17:01 補充： (1+i)z^2－(2+3i)z+2+i=0 z=((2+3i)+/-3i)/(2+2i) (1) z=(2+3i+3i)/(2+2i) =(1+3i)/(1+i) =(1+3i)(1－i)/[(1+i)(1－i)] =(1－i+3i+3)/2 =2+i (2) z=((2+3i)－3i)/(2+2i) =2/(2+2i) =1/(1+i) =(1－i)/2
其他解答:
(1+i)z^2-(2+3i)+2+i=0 (1+i)z^2 -2 -3i +2 +i=0 (1+i)z^2 -3i +i = 0 (1+i)z^2 -2i = 0 (1+i)z^2 = +2i z^2 = +2i/(1 + i) ? ? ? ? ?2i(1 – i) z^2 = --------------- ? ? ? ? (1+i)(1 – i) ? ? ? ? ?2i(1 – i) z^2 = ------------ ?(Note: i^2 = -1) ? ? ? ?(1– i^2) ? ? ? ? ?2i(1 – i) z^2 = ------------ ? ? ? ? ? 1–(-1) ? ? ? ? ?2i(1 – i) z^2 = ------------ ? ? ? ? ? ? ?2 z^2 = i – i^2 z^2 = i – (-1) z^2 = 1 + i ? Taking square root on both sides z = square root of (1 + i) ?or (1 + i)^0.5|||||好明顯有問題 你話無打錯, 都get 唔到問乜 -(2+3i)+2+i 點解會咁寫? 麻煩你寫清楚d 2010-08-25 22:53:45 補充： Z cannot be found. Z can only be determined in terms of i|||||Have you missed z next to (2+3i)? 2010-08-25 17:56:01 補充： If question isn't wrong, then you can solve for z simply by rearranging the real terms and imaginary terms, and take the square root of both side to get z,
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